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Which force is the centripetal one in a banked curve?

  1. Feb 24, 2016 #1
    Hello Forum,

    In the situation of a car banking a curve there are two forces: weight W and the normal force N. Theta is the banking angle between the road and the horizontal.

    The vector addition between W and N gives a net force F_net that is directed toward the center. That F_net is the centripetal force. Both forces take part in generating the centripetal force.

    If we orient the Cartesian axes so that the x-axis is point toward the center of the we discover than only the contact force N has a component along that axis and generates the centripetal force...

    How should I reconcile this difference?

    From the vector diagram, we see both forces participating in creating the centripetal force while in the component analysis we only see the normal force participating. However, if the x-axis was directed along the slope, then both forces W and N would have component along the direction toward the center of curvature and it would turn out that both forces participate to the net force also from a component analysis point of view...

    thanks,
    fog37
     
  2. jcsd
  3. Feb 24, 2016 #2

    rcgldr

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    There's also a friction force between the tires and the banked curved. Assuming a level turn, then gravity is not a component of centripetal force. The centripetal force is the sum of the horizontal components of the normal and friction forces.
     
  4. Feb 24, 2016 #3
    True. I am assuming no friction for now and the car moving at the "right" speed based on the banking angle.

    On the flat level, friction is the force that plays the role of the centripetal force. Friction depends on the normal force which depends on weight
     
  5. Feb 24, 2016 #4
    Lacking friction, the normal force (or part of it) is centripetal.

    The centripetal force is always normal to the path of motion and toward the center.

    Remember a centripetal force is unopposed. It causes acceleration toward the center.

    Thus in the situation you described, part of the normal force will exactly counteract gravity. That's the part of the vector pointing up. Another part of the vector points to the center. That is the component that is centripetal.
     
  6. Feb 25, 2016 #5

    A.T.

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    Not all summed vectors must have a component parallel the vector sum. Your misunderstanding is about maths, not physics.
     
  7. Feb 25, 2016 #6
    I agree with all of you. I may have not been clear. I am just pointing out that from the force component analysis only the normal force seems to be responsible for the net centripetal force. But for the vector addition, it is clear that the net centripetal force is due to both the normal force and the weight force.

    The FBD, depending on how we orient the Cartesian axes can give a conceptual picture that is different than the actual vector picture. for example, consider the x-axis aligned with the slope of the curve. You will now state that both weight and normal force have a component along the direction that goes to the center of the curve and the sum of those two components is the centripetal force. The way we orient the coordinate axes is arbitrary....
     
  8. Feb 25, 2016 #7

    A.T.

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    Whether weight has component towards the center or not does not depend on the choice of coordinates.
     
  9. Feb 25, 2016 #8
    Yes, that is exactly my point.

    So, it seems cannot uniquely state that the centripetal force is due to the normal force only because by changing the coordinate system direction we can state that both the normal and the weight forces participate to the centripetal force...
     
  10. Feb 25, 2016 #9

    A.T.

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    No. It's the opposite of what you state.
     
  11. Feb 25, 2016 #10
    The normal force is normal (perpendicular) to the surface. Gravity is straight down. Since there's a slope to the bank curve, the normal force is not straight up. It has an up component and a "to the center" component. Lacking a flying car, the up component exactly matches the force of gravity.

    I should add that I assumed the circular track is level in the sense that the center is at the same height as the car. Were the track canted (like a tilt-a-wheel), it would be more complicated.
     
  12. Feb 25, 2016 #11
    For a stationary vehicle on a level track, the entire weight of the vehicle is normal to the track surface. As the track bank is increased, the normal force against the pavement decreases and a component of tire friction against the banked track begins to offset some fraction of the vehicle's weight against gravity. Keep increasing the bank, and either the tires will lose their grip and allow the vehicle to slide sideways, or if the tires are super "sticky" (or the vehicle has a very high center of gravity), the vehicle will topple.

    But your assumption that the car is moving at the "right" speed on the banked (and, I assume, also appropriately curved) track means that the track must be applying an increased normal force which is greater than the normal force for a similarly banked, but NOT curved track. The horizontal component of that increased normal force is directed back to the center of the curve.

    It's really hard to eliminate the influence of tire friction in such an example. If the car is traveling at less than the "right" speed, tire friction is holding the car from sliding down (towards the center of curve), while a car traveling in excess of the "right" speed has tire friction fighting the car from sliding up the incline, away from the center of the curve. The tire-friction contribution in such a dynamic is not negligible.
     
    Last edited: Feb 25, 2016
  13. Feb 25, 2016 #12
    This is what I was trying to explain:

    upload_2016-2-25_22-26-52.png
     
  14. Feb 25, 2016 #13
    For a banked curve, your components diagram should show the apparent weight of the car normal to the track surface, not vertical. The components should sum to zero. The track "feels" the weight of the turning car as the opposite of the vector sum of the vertical and horizontal components depicted.
     
  15. Feb 25, 2016 #14
    In a free body diagram there is no apparent weight. The only two forces are W and F_N which have their x and y components.
     
  16. Feb 26, 2016 #15
    Let's back up here a second... The centripetal force is not caused by the angle of the track -- a straight track could have a similar slope, but it would not cause the car to travel in a circular path (although the driver would have to fight gravity trying to pull him down the slope, that would not be a centripetal force). The centripetal force here is the result of the curved track continuously changing the direction of travel. The centripetal force manifests itself in the continuously changing velocity vector as the car circles the track (if the circular track were flat, the lateral friction of the tires would have to provide the direction change in place of the slope of the track).
     
    Last edited: Feb 26, 2016
  17. Feb 26, 2016 #16
    Sure, I agree.

    I guess my point is: what force(s) determine the centripetal acceleration? Both W and F_N or only F_N? According to the vector diagram both forces but according to the component diagram only F_N
     
  18. Feb 26, 2016 #17

    A.T.

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    Depends on the definition of "determine", which hints that it's an irrelevant question to ask.
     
  19. Feb 26, 2016 #18
    I think the vector equation is what represents best the physical aspect of the problem. In the sense that both forces are needed to provide the centripetal force in the sense that the normal force is also dependent on Weight.

    The component diagram can make things look different. Sometimes the choice of a certain coordinate system completely can make equal to zero the components of a vector....
     
  20. Feb 26, 2016 #19

    A.T.

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    I wouldn't concentrate on physical aspects here, because the issue is purely mathematical about how vectors work.

    The centripetal net force is simply the vector sum of both forces, so it obviously is affected by both. The dependency between them is completely irrelevant to this.

    Two orthogonal vectors will stay orthogonal, no matter how you rotate your system. They will never have a component parallel the other.
     
  21. Feb 27, 2016 #20
    With the car in motion, the stated problem is a dynamic, not a static one. A vehicle traveling around a circular track at a constant speed is still accelerating by virtue of the continuously changing direction of the constant speed's instantaneous vector as the car moves around the track. The centripetal acceleration is the rate of change of the velocity vector at that "right speed" you mentioned earlier. The "right speed" depends upon both the bank angle and the track radius. For any specific track, it is that speed for which the tires don't contribute any lateral forces to keep the car at a constant distance from the center of the track circle (that's when the banked track "feels" the apparent weight of the vehicle normal to the track surface -- that apparent weight is the normal component of the weight of the car, plus the normal component of the centripetal force).

    If the vehicle is traveling at some speed other than the "right speed", then the contribution of tire friction against side slip doesn't change the centripetal force, but must add to (or subtract from) the centripetal force if the car is to stay in it's intended "lane".
     
    Last edited: Feb 27, 2016
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